The value of $${\left( {\frac{{1 - \cot \theta }}{{1 - \tan \theta }}} \right)^2} + 1,$$    0° < θ < 90°, is equal to:

Solution(By Examveda Team)

$$\eqalign{ & {\left( {\frac{{1 - \cot \theta }}{{1 - \tan \theta }}} \right)^2} + 1 \cr & = {\left( {\frac{{1 - \frac{{\cos \theta }}{{\sin \theta }}}}{{1 + \frac{{\sin \theta }}{{\cos \theta }}}}} \right)^2} + 1 \cr & = {\left( {\frac{{\sin \theta - \cos \theta }}{{\sin \theta }} \times \frac{{\cos \theta }}{{\cos \theta - \sin \theta }}} \right)^2} + 1 \cr & = {\left( {\frac{{\sin \theta - \cos \theta \times \cos \theta }}{{ - \sin \theta \left( {\sin \theta - \cos \theta } \right)}}} \right)^2} + 1 \cr & = {\left( {\frac{{ - \cos \theta }}{{\sin \theta }}} \right)^2} + 1 \cr & = {\cot ^2}\theta + 1 \cr & = {\text{cose}}{{\text{c}}^2}\theta \cr} $$